# How do you factor #64x^3+8=0#?

##### 1 Answer

Mar 7, 2016

#### Explanation:

This is a sum of cubes, since both

Differences of cubes factor as follows:

#a^3+b^3=(a+b)(a^2-ab+b^2)#

In

#64x^3+8=0#

#(4x+2)((4x)^2-4x(2)+(2)^2)=0#

#(4x+2)(16x^2-8x+4)=0#

#8(2x+1)(4x^2-2x+1)=0#